Integral Calculus Question 72
Question: $ \int_{{}}^{{}}{\frac{x^{2}-1}{x^{4}+x^{2}+1}\ dx=} $
[AISSE 1990]
Options:
A) $ \frac{1}{2}\log ( \frac{x^{2}+x+1}{x^{2}-x+1} )+c $
B) $ \frac{1}{2}\log ( \frac{x^{2}-x-1}{x^{2}+x+1} )+c $
C) $ \log ( \frac{x^{2}-x+1}{x^{2}+x+1} )+c $
D) $ \frac{1}{2}\log ( \frac{x^{2}-x+1}{x^{2}+x+1} )+c $
Show Answer
Answer:
Correct Answer: D
Solution:
The given function can be written as $ \int_{{}}^{{}}{\frac{( 1-\frac{1}{x^{2}} )}{{{( x+\frac{1}{x} )}^{2}}-1}},dx $ Put $ x+\frac{1}{x}=t\Rightarrow ( 1-\frac{1}{x^{2}} ),dx=dt, $ then it reduces to $ \int_{{}}^{{}}{\frac{dt}{t^{2}-1}}=\frac{1}{2}\log | \frac{t-1}{t+1} |+c $ $ =\frac{1}{2}\log ( \frac{x+\frac{1}{x}-1}{x+\frac{1}{x}+1} )+c=\frac{1}{2}\log ( \frac{x^{2}-x+1}{x^{2}+x+1} )+c. $
BETA
